JEE Advanced 2020 · Paper 2, Mathematics question 9

Value at 0. $f(0)=0\cdot g(0)=0$.

Derivative.
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{f(h)\,(1+f(x))}h=(1+f(x))\lim_{h\to0}g(h)=1+f(x).$$
So $f$ is differentiable everywhere, and (A) is true.

(D) $f'(0)=1+f(0)=1$. True.

(C) Solving $f'=1+f$ with $f(0)=0$ gives $f(x)=e^x-1$. Then $f'(1)=e\ne1$. False.

(B) $g(x)=\dfrac{e^x-1}x$ for $x\ne0$, and $g(0)=1$ makes $g$ the sum of the power series $\sum\frac{x^n}{(n+1)!}$, which is differentiable everywhere. True.

(A), (B), (D)